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3 years ago

Beryllium has a charge of +2, and bromine has a charge of –1. Which is the best name for the ionic bond that forms between them?

1 answer:

6 0

There are 2 elements:
Beryllium ( Be + 2 ), metal
Bromine ( Br - 1 ), non - metal
They form the ionic bond:
Be + Br2 → Be Br2
Naming the ionic compound:
on the 1st place is metal. On the 2nd place is non-metal. If compound is binary, use the element name and change ending to - ide .
Answer:
Beryllium Bromide.

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If a force of 10 N acts on an object and an additional force of 6 N acts on the object in

PIT_PIT [208]

Answer:

Explanation:

For this kind of problem, forces add. F = F1 + F2

F1 = 6 N

F2 = 10 N

F = 6N + 10N

F = 16N

6 0

2 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0

3 years ago

An atom with 9 protons, 11 electrons, and 10 neutrons has a net charge of

LiRa [457]

You find net charge by subtracting the number of electrons from the number of protons (since protons are positive and electrons are negative). 9 - 10 = -1

5 0

3 years ago

You could swallow a capsule of germanium, Ge (atomic number 32), without significant ill effects. If a proton were added to each

Ronch [10]

Answer: A

Explanation:

Germanium, is atomic number 32, and if you add a proton, Germanium converts to arsenic. arsenic is very poisonous.

4 0

3 years ago

A proton enters a region of constant magnetic field, perpendicular to the field and after being accelerated from rest by an elec

kicyunya [14]

Answer:

B = 1.1413 10⁻² T

Explanation:

We use energy concepts to calculate the proton velocity

starting point. When entering the electric field

Em₀ = U = q V

final point. Right out of the electric field

em_f = K = ½ m v²

energy is conserved

Em₀ = Em_f

q V = ½ m v²

v = $\sqrt{2qV/m}$

we calculate

v = $\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }$

v = $\sqrt{632.3353 \ 10^8}$

v = 25.15 10⁴ m / s

now enters the region with magnetic field, so it is subjected to a magnetic force

F = m a

the force is

F = q v x B

as the velocity is perpendicular to the magnetic field

F = q v B

acceleration is centripetal

a = v² / r

we substitute

qvB =1/2 m v² / r

B = v $\frac{m v}{2 q r}$

we calculate

B = $\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}$

B = 1.1413 10⁻² T

8 0

3 years ago